I need help answering this question in relation to Stoichiometry volume-mass in my chemistry class.

When an electric field is applied to a metallic crystal, the movement of electrons is .

Talja [164]

Your answer is random because metals form huge structures in which electrons in the outer shell of the metal it makes atoms free to move.

8 0

3 years ago

Read 2 more answers

2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2

miv72 [106K]

<h3>Answer:</h3>

4 g AgCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given] 5.0 g AgNO₃

<u>Step 2: Identify Conversions</u>

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})$
Multiply/Divide: $\displaystyle 4.21533 \ g \ AgCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

4.21533 g AgCl ≈ 4 g AgCl

3 0

3 years ago

Which distinguishes an atom of one element from an atoms of a different element?

dalvyx [7]

Answer:

The number of protons you welcome

Explanation:

7 0

3 years ago

9. Let's say, you need to make 2.00 L of 0.05 M copper (II) nitrate solution. How many grams of

Vlad [161]

Answer:

18.76 g of copper II nitrate

Explanation:

Now recall that we must use the formula;

n= CV

Where;

n= number of moles of copper II nitrate solid

C= concentration of copper II nitrate solution

V= volume of copper II nitrate solution

Note that;

n= m/M

Where;

m= mass of solid copper II nitrate

M= molar mass of copper II nitrate

Thus;

m/M= CV

C= 0.05 M

V= 2.00 L

M= 187.56 g/mol

m= the unknown

Substituting values;

m/ 187.56 g/mol = 0.05 M × 2.00 L

m= 0.05 M × 2.00 L × 187.56 g/mol

m= 18.76 g of copper II nitrate

Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.

6 0

3 years ago

0. Give an example of how an organism can use surface tension.<br><br><br>

Lady_Fox [76]

Answer:

In case of plants surface tension help to support the transpiration pull.

Explanation:

Surface tension is a force that hold the molecules in the surface to minimize the surface area.

During evaporation of excess amount of water from the stomata of leaves of plants by transpiration a surface tension is generated.

The generated surface tension helps to maintain the water column within the xylem tissue by the absorption of water from the soil by the roots.

3 0

4 years ago